Question 653987: the GCF of 3240, 3600 and an unknown third number is 2^2 X 3^2 ;
the LCM of the three numbers is 2^4 X 3^5 X 5^2 X 7.
Find the third number...
First, write 3240 and 3600 as a product of prime factors
3240 = _______
3600= ________
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
I'll show you how to do one of these and then you can do the other for yourself:
3240 is an even number, therefore divisible by 2. 3240 divided by 2 is
1620 is an even number, therefore divisible by 2, 1620 divided by 2 is
810 is an even number, therefore divisible by 2, 810 divided by 2 is
405 is an odd number, not divisible by 2. The sum of the digits of 405 is 9 which is divisible by 3, hence 405 is divisible by 3. 405 divided by 3 is
135 odd, sum of digits is 9, divisible by 3. 135 divided by 3 is
45 odd, sum of digits is 9, divisible by 3. 45 divided by 3 is
15 odd, sum of digits is 6, divisible by 3. 15 divided by 3 is
5 which is prime.
Summary: 
Test for divisibility by 5: divisible by 5 if ends in 5 or 0.
All other trial divisors, just do by trial and error.
Always start with 2 and work up. If you get past the square root of the number you are trying to factor without finding a prime that divides it evenly, then the starting number is prime.
Example:
Test 317. Square root is 17.8 and change.
Not even, so not divisible by 2. Sum of digits is 11, 11 not divisible by 3, so 317 not divisible by 3. Does not end in 5, so not divisible by 5. Try 7, no. Try 11, no. Try 13, no. Try 17, no. Next prime number exceeds the square root of 317, hence 317 is prime.
John
My calculator said it, I believe it, that settles it
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